A) \[\sqrt{2gl}\]
B) \[\sqrt{3gl}\]
C) \[\sqrt{gl/2}\]
D) \[\sqrt{gl/3}\]
Correct Answer: D
Solution :
[d] \[T-mg\cos \theta =\frac{m{{v}^{2}}}{R}\] Given \[T=mg\] \[mg-mg\cos \theta =\frac{m{{v}^{2}}}{R}\] \[g(1-cos\theta )=\frac{{{v}^{2}}}{R}\] C.O.M.E. at A and B; \[\Delta K+\Delta U=0\] \[\left( \frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}} \right)+mg(R-Rcos\theta )=0\] \[\Rightarrow {{v}^{2}}-{{u}^{2}}=-2gR(1-cos\theta )\] \[\Rightarrow {{v}^{2}}-{{(\sqrt{gl})}^{2}}=-2{{v}^{2}}\] \[3{{v}^{2}}=gl\Rightarrow v=\sqrt{\frac{gl}{3}}\]You need to login to perform this action.
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